Question

ГЕОМЕТРИЯ, 11 КЛАСС, 56 БАЛЛОВ В треугольной пирамиде ABCD известно: AB=CD=4, AD=BC=8, ∠ABC=120∘. Найдите R^2, где R — радиус наименьшего шара, в который можно поместить такую пирамиду. С простым или хотя-бы поверхностным пояснением, спасибо

Answer 1

Ответ:

R² = 28 ед.

Объяснение:

По теореме косинусов АС² = АВ² + ВС² - 2·АВ·ВС·Cos120.

АС² = 16 + 64 - 2·4·8·(-1/2) = 112. √112 ≈  10,6 ед.

Ребро АС - большее из всех ребер пирамиды (по теореме о неравенстве треугольника).

Найдем высоту треугольника ADC h.

Она делит треугольник на два прямоугольных треугольника. Выразим в обоих треугольниках h по теореме Пифагора и приравняем оба выражения:

AD² - (AC-x)² = DC² - x².

64 - 112 + (2√112)·x - x² = 16 - x². =>  x ≈ 3.

h ≈ √(16-9) ≈ 2,6 ед. Следовательно, высота нашей пирамиды еще меньше, так как перпендикуляр меньше наклонной из точки D к плоскости основания.

Поэтому минимальный диаметр шара, в который поместится данная пирамида, равен АС, поскольку диаметр - наибольшая из хорд.

AC = D = 2R.

4R² = 112  => R² = 28 ед.

Answer 2

Problem Analysis

We are given a triangular pyramid ABCD, where AB = CD = 4, AD = BC = 8, and ∠ABC = 120°. We need to find the square of the radius of the smallest sphere that can contain this pyramid.

Solution

To find the square of the radius of the smallest sphere, we need to find the height of the pyramid and the radius of the inscribed sphere.

# Finding the Height of the Pyramid

To find the height of the pyramid, we can use the Pythagorean theorem. Let's consider triangle ABC. Since AB = CD = 4 and ∠ABC = 120°, we can use the Law of Cosines to find the length of BC.

Using the Law of Cosines: BC^2 = AB^2 + AC^2 - 2 * AB * AC * cos(∠ABC)

Substituting the given values: BC^2 = 4^2 + AC^2 - 2 * 4 * AC * cos(120°)

Simplifying: BC^2 = 16 + AC^2 + 8 * AC

Since AD = BC = 8, we can substitute BC with 8: 8^2 = 16 + AC^2 + 8 * AC

Simplifying further: 64 = 16 + AC^2 + 8 * AC

Rearranging the equation: AC^2 + 8 * AC - 48 = 0

Now we can solve this quadratic equation to find the value of AC.

# Solving the Quadratic Equation

Using the quadratic formula: AC = (-b ± √(b^2 - 4ac)) / (2a)

Substituting the values: AC = (-(8) ± √((8)^2 - 4(1)(-48))) / (2(1))

Simplifying: AC = (-8 ± √(64 + 192)) / 2 AC = (-8 ± √256) / 2 AC = (-8 ± 16) / 2

We have two possible values for AC: AC1 = (-8 + 16) / 2 = 4 AC2 = (-8 - 16) / 2 = -12

Since AC represents a length, we can discard the negative value. Therefore, AC = 4.

Now, we can substitute the value of AC back into the equation BC^2 = 16 + AC^2 + 8 * AC to find the length of BC.

BC^2 = 16 + 4^2 + 8 * 4 BC^2 = 16 + 16 + 32 BC^2 = 64

Taking the square root of both sides, we find that BC = 8.

# Finding the Radius of the Inscribed Sphere

To find the radius of the inscribed sphere, we need to find the height of the pyramid. Since the pyramid is equilateral, the height is given by the formula:

Height = (sqrt(3) / 2) * side length

In this case, the side length is BC = 8. Substituting the value, we find:

Height = (sqrt(3) / 2) * 8 = 4 * sqrt(3)

Now, we can find the radius of the inscribed sphere using the formula:

Radius = (1/3) * Height

Substituting the value of Height, we find:

Radius = (1/3) * (4 * sqrt(3)) = (4/3) * sqrt(3)

Finally, we need to find the square of the radius:

R^2 = [(4/3) * sqrt(3)]^2 = (16/9) * 3 = 16/3.

Therefore, the square of the radius of the smallest sphere that can contain the given pyramid is 16/3.

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